$$
\sqrt{(x-x')^2+(y-y')^2+z^2} \approx \sqrt{x^2+y^2+z^2-2xx'-2yy'}
$$
where the terms square in ##x'## and ##y'## have been omitted. Then make substitution ##r^2 = x^2+y^2+z^2##,
$$
r\sqrt{1-2\frac{x}{r^2}x'-2\frac{y}{r^2}y'}
$$
and apply Taylor expansion truncating the third term.

$$
\sqrt{(x-x')^2+(y-y')^2+z^2} \approx \sqrt{x^2+y^2+z^2-2xx'-2yy'}
$$
where the terms square in ##x'## and ##y'## have been omitted. Then make substitution ##r^2 = x^2+y^2+z^2##,
$$
r\sqrt{1-2\frac{x}{r^2}x'-2\frac{y}{r^2}y'}
$$
and apply Taylor expansion truncating the third term.

Hello, I am really sorry but could you provide me the taylor expansion of it?

To simplify the appearance, you can make the substitution ##-2\frac{x}{r^2}x'-2\frac{y}{r^2}y' = u## so that
$$
r\sqrt{1-2\frac{x}{r^2}x'-2\frac{y}{r^2}y'} = r(1+u)^p
$$
where ##p=1/2##. Now look up online or in your textbook examples of Taylor series, especially the series which corresponds to a form ##(1+u)^p## with ##|u|<1## as is the case here.

To simplify the appearance, you can make the substitution ##-2\frac{x}{r^2}x'-2\frac{y}{r^2}y' = u## so that
$$
r\sqrt{1-2\frac{x}{r^2}x'-2\frac{y}{r^2}y'} = r(1+u)^p
$$
where ##p=1/2##. Now look up online or in your textbook examples of Taylor series, especially the series which corresponds to a form ##(1+u)^p## with ##|u|<1## as is the case here.